AQ1.9 Q-6

[ajaykuma...@gmail.com]

Sir, pls explain question no.6 of AQ 1.9.

[Debajyoti Biswas]

This question asks whether the given function is bijective. Now we know that a function is bijective only when it is both injective and surjective.  Now put A=0 or B=0 or B={-1,1} as per the options given and for each case check whether they are injective and surjective.

[Sree Kavitha]

stilldonot understand

[Debajyoti Biswas]

For checking whether bijective, the function is to be injective and surjective. If A=0, f(1)=f(2)=B (for example). Hence this is not injective. Hence not bijective.

Next, for A={-1,1}, f(x) is becoming injective and for every element in the codomain of f, there is at least one element in the domain of f such that f(x) = y. So this is surjective.